 Hi everyone, we are now at the end of the end. We are at the last part of the last lecture of our calculus one series. And so before we put this course to bed, I do want us to say a little bit more about integration and in particular with its relation to symmetric functions. There's two types of symmetries that we've been very interested in this semester. There's the idea of even symmetry and odd symmetry. And do recall that a function is even when if we take f of negative x, this is equal to just f of x. So if you reflect the function across the y-axis, you get the exact same graph. It's unaffected by that reflection. Odd functions are similar here. They have the property that if you take f of negative x, this is equal to negative f of x. So geometrically speaking, a function is odd if a reflection across the y-axis is equivalent to a reflection across the x-axis. And that's equivalent to saying that the, if you rotate around the origin, you get the 180 degrees, you get the exact same picture here. And so if your function's even, I want you to notice that if you integrate an even function from negative a to a, f of x dx, this will just equal two times the integral from a to zero of f of x dx. And the basic idea behind that is the following. If your function is even, then maybe it looks something like the following. It's symmetric with respect to the y-axis. So if you go from negative a to a, it's the exact same distance from the y-axis, then the area under the curve here is just gonna be, the total area of the curve is just twice as mount as the area under the curve from the y-axis, just to the right. That is the right hand side there. So if we just wanna go from zero to the right, that helps us out here a lot. And why does it help us out a lot? Well, it really just comes down to benefits of arithmetic, right? We have to calculate these things. There's some number crunching going on there. If I had a choice between any number or zero, hands down, I'm gonna wanna pick zero. And so that bit of arithmetic can help us out here in this type of calculation. When it comes to an odd function on the other hand, though, how does odd symmetry affect the anti-derivative? I should say, how does it affect the integral? Well, if you integrate from negative a to a of f of x dx, the area under the curve is just zero. Absolutely zero, which is just fantastic. And so geometrically, what we're saying is the following. If we have some type of, some type of, let's take an odd function, so it's symmetric with respect to the y-axis. And let's say we go from negative a over here to a. We're saying that the amount of area that's above the x-axis here is gonna be opposite and equal to the amount of area that's below the x-axis. So when you take the net sum of these things, it always adds up to be zero. Which one nice thing about odd functions here is if you integrate an odd function along the symmetric interval, it does need to be a continuous odd function, by the way. But if you integrate a continuous odd function on a symmetric integral interval, then the area under the curve is always zero. You don't need an anti-derivative. You can recognize symmetry and then to say it's zero, drop the mic and walk away. And the basic reasons behind this, so I've given you some geometric intuition behind that, but we can prove this thing algebraically as well. Because if you were integrating negative a to a of f of x dx, in both situations, you could break this up to negative a to zero f of x dx and then you add that to the integral of zero to a f of x dx. It's always true that you can take an integral and break it up into pieces using some intermediate value, some zero that sits between negative a and a right here. And so for the first case, if we have, well, and so I guess for both cases still let's switch the order of the first one right here, switch the zero and the negative a. And so you're gonna get this statement, the integral of zero a f of x dx. I just bought the second one first. And then you can subtract from that the integral from zero to a f of x dx, like so, and so this original expression is just equal to this right here. And so now this is where the even and odd business is gonna come into play. So let's suppose that first that we are even, right? And if you're an even function, that means that f of x, f of negative x equals f of x. And so what we're gonna do is we're gonna do a U substitution here, just for the second one here. There should be a minus sign right there, I forgot to put. So for the second one, we're gonna use the substitution U equals negative x and hence dx, sorry, du is equal to just negative dx. And so this puts our integral looking as, we have the integral from zero to a f of x dx plus the integral from zero to a of f of negative x. So I should say f of U, what did I do here? I'm sorry, I do need a negative U right there. So you get f of U du. So the idea is when you make the substitution x equals, x equals U equals negative x. This is the same thing, same negative U equals x right here. So you're gonna get this f of negative U going on right there. This also will change the upper bound. So it'll be a A instead of a negative A right there. And so at this moment's where you can use symmetry. So if your function was even, then the substitution here is that f of negative U will become f of U. And so under the even hypothesis, this statement looks like the integral from zero to A f of x dx plus the integral of zero to A of f of U du. And although the variables are different, f of x and f of U, it's the same thing. And this is just gonna double up and give us two times the integral from zero to A of f of x dx, okay? That's what happens for the even case. For the odd case, though, for the odd case, things are a little bit differently. You get zero to A f of x dx, and then you get plus the integral from zero to A. Well, in the odd case, f of negative x here notices negative f of x. So when you have this f of negative U, this becomes negative f of U du. And you can move this negative sign out in front. And so you're gonna get the integral from zero to A of f of x dx minus integral from zero to A of f of U du. And so again, even though the variable is different, x versus U, this is the exact same quantity and these things will cancel out to give us a zero there. So that's enough talking about the justification. Let's actually see how one uses this in practice. So this function, or this integral right here, integrate from negative two to two, x to the sixth plus one dx. And so you'll notice here that we have a symmetric interval. We're going from negative two to two. So when I see things like that, I sort of ask myself, is there any type of symmetry going on here? And symmetry is not so hard to check here. If you take f of x to be x to the sixth plus one, then f of negative x will equal negative x to the sixth plus one. And when you take an even power of a negative, that actually makes it a positive. We get x to the sixth plus one, which is just f of x. This is indicative that our function is even. And the reason we call it even functions, if you look at all the powers of x, you have only even power, six is even number. The one of course is x to the zero, which is an even power as well. So this is an even function. So by the integral of even symmetric functions we saw before, this will equal two times the integral from zero to two of x to the sixth plus one dx. We do still have to calculate an anti-derivative with that anti-derivative will look like x to the seventh over seven plus x, but we evaluate from zero to two. And so when we plug in the two, we get two times, two to the seventh over seven plus two. But then we're gonna get, when we plug in the zero, we get zero to the seventh over seven plus zero. You'll notice that this last term right here, everything just goes to zero. And that's exactly what's gonna happen right here. And so this observation of symmetry is just for the arithmetic benefit of things. Plug it in zero with all these fractions is so much nicer than any other number. So we use that as a benefit just to simplify the calculation of the integral we have in front of us. The rest of what we still have to do, we start doing our powers of two. We have to do two to the seventh. We can use a calculator, but if we've played the game 2048 enough, then we might actually have these memorized. And so two to the seventh, we can just count on our fingers if we have to. We get two, four, eight, 1632, 64, 128 over seven plus four, two times two. Of course, we can times two by another, or 128 by another two. So we get 256 over seven. And we have to add four to that, which we're gonna write that as 28 over seven. Combining like terms, we end up with the final form 284 over seven. So it's not that we could avoid all of the arithmetic, but the fact that we were able to plug in zeros at one point did make the final result a little bit easier to do. Again, it's not necessarily a saving grace here, but it makes it a lot easier to do with this one. We turns out in calculus two, one sees this type of even symmetry all the time and we're very grateful for it. What about this one right here? The integral from negative one to one of tangent of x over one plus x squared over x to the fourth. Look at this one right here. It's a symmetric interval again. So is it symmetric? Is the function symmetric? And so if we take f of x to be tangent over one plus x squared plus x to the fourth, this one might not be as obvious here, but if you look at f of negative x, oh boy. f of negative x there. You're gonna end up with tangent of negative x and you're gonna have one over negative x squared plus negative x to the fourth. And so tangent is an odd function. So you get negative tangent of x on the top. On the bottom, you have even. All the even powers absorb the negative signs. And so you'll notice you end up just negative f of x from negative f of x there. So this indicates that the original integral is odd. And so if we wanna calculate an antiderivative of this thing, that is gonna be monstrous. I will be impressed if you can find an antiderivative of this function at this stage of the game, right? But we don't need an antiderivative because we're integrating an odd function along the symmetric interval. The theorem before says that this thing is zero and we're done. That's all that one has to do. As a fun little story, a couple of years ago, I gave a question very similar to this on a final for calculus one. It was a multiple choice question, right? And all the students had to do was select the option zero. That's all I wanted to do. I thought I didn't think it was gonna be such a hard problem but it blew their minds away. They did some crazy stuff trying to find antiderivatives. And this is calculus one, mind you, they had no idea what to do. So I now put this as a warning, right? Look out for odd functions. If you can use those to evaluate antiderivatives, sorry, if you can use symmetry to help you with integrals, you might be able to avoid having to compute antiderivatives, which is the fundamental hardest part of these type of questions here. And so that brings us to the end of this video here and also brings us to the end of our lecture series, Math 1210 Calculus One for students at Southern Utah University. It's been a good course. I hope everyone enjoyed these videos and learned a whole lot. If you have any questions, feel free to always post them in the comments below. Like these videos, subscribe so you can learn more about fun math things in the future, right? Check out my playlist for Calculus Two, Math 1220. If you wanna learn some more about calculus, all right? And I hope to see you all sometime in the future. Everyone keep on calculating, bye.